Divide the following complex numbers. $ \dfrac{3+9i}{-2-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2+i}$ $ \dfrac{3+9i}{-2-i} = \dfrac{3+9i}{-2-i} \cdot \dfrac{{-2+i}}{{-2+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(3+9i) \cdot (-2+i)} {(-2-i) \cdot (-2+i)} = \dfrac{(3+9i) \cdot (-2+i)} {(-2)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(3+9i) \cdot (-2+i)} {(-2)^2 - (-1i)^2} = $ $ \dfrac{(3+9i) \cdot (-2+i)} {4 + 1} = $ $ \dfrac{(3+9i) \cdot (-2+i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({3+9i}) \cdot ({-2+i})} {5} = $ $ \dfrac{{3} \cdot {(-2)} + {9} \cdot {(-2) i} + {3} \cdot {1 i} + {9} \cdot {1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-6 - 18i + 3i + 9 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-6 - 18i + 3i - 9} {5} = \dfrac{-15 - 15i} {5} = -3-3i $